Barron's SAT Subject Test Math Level 2, 10th Edition (2012)
Part 2. REVIEW OF MAJOR TOPICS
Chapter 3. Numbers and Operations
3.1 Counting
VENN DIAGRAMS
Counting problems usually begin with the phrase “How many . . .” or the phrase “In how many ways . . .” Illustrating counting techniques by example is best.
EXAMPLES
1. A certain sports club has 50 members. Of these, 35 golf, 30 hunt, and 18 do both. How many club members do neither?
Add 35 and 30, then subtract the 18 that were counted twice. This makes 47 who golf, hunt, or do both. Therefore, only 3 (50 – 47) do neither.
2. Among the seniors at a small high school, 80 take math, 41 take Spanish, and 54 take physics. Ten seniors take math and Spanish; 19 take math and physics; and 12 take physics and Spanish. Seven seniors take all three. How many seniors take math but not Spanish or physics?
A Venn diagram will help you sort out this complicated-sounding problem.
Start with the 7 who take all three courses. Since 10 take math and Spanish, this leaves 3 who take math and Spanish but not physics. Use similar reasoning to see that 5 take physics and Spanish but not math, and 12 take math and physics but not Spanish.
Finally, using the totals for how many students take each course, we can conclude that 58 (80 – 3 – 7 – 12) take math but not physics or Spanish.
EXERCISE
1. There are 50 people in a room. Twenty-eight are male, and 32 are under the age of 30. Twelve are males under the age of 30. How many women over the age of 30 are in the group?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
MULTIPLICATION RULE
Many other counting problems use the multiplication principle.
EXAMPLES
1. Suppose you have 5 shirts, 4 pairs of pants, and 9 ties. How many outfits can be made consisting of a shirt, a pair of pants, and a tie?
For each of the 5 shirts, you can wear 4 pairs of pants, so there are 5 
4 = 20 shirt-pants combinations. For each of these 20 shirt-pants combinations, there are 9 ties, so there are 20 9 = 180 shirt-pants-tie combinations.
2. Six very good friends decide they will have lunch together every day. In how many different ways can they line up in the lunch line?
Any one of the 6 could be first in line. For each person who is first, there are 5 who could be second. This means there are 30 (6 5) ways of choosing the first two people. For each of these 30 ways, there are 4 ways of choosing the third person. This makes 120 (30 4) ways of choosing the first 3 people. Continuing in this fashion, there are 6 5 4 3 2 1 = 720 ways these 6 friends can stand in the cafeteria line. (This means that if they all have perfect attendance for 4 years of high school, they could stand in line in a different order every day, because 720 = 4 180.)
3. The math team at East High has 20 members. They want to choose a president, vice president, and treasurer. In how many ways can this be done?
Any one of the 20 members could be president. For each choice, there are 19 who could be vice president. For each of these 380 (20 19) ways of choosing a president and a vice president, there are 18 choices for treasurer. Therefore, there are 20 19 18 = 6840 ways of choosing these three club officers.
4. The student council at West High has 20 members. They want to select a committee of 3 to work with the school administration on policy matters affecting students directly. How many committees of 3 students are possible?
This problem is similar to example 3, so we will start with the fact that if they were electing 3 officers, the student council would be able to do this in 6840 ways. However, it does not matter whether member A is president, B is vice president, and C is treasurer or some other arrangement, as long as all 3 are on the committee. Therefore, we can divide 6840 by the number of ways the 3 students selected could be president, vice president, and treasurer. This latter number is 3 2 1 = 6, so there are 1140 (6840 ÷ 6) committees of 3.
EXERCISES
1. M & M plain candies come in six colors: brown, green, orange, red, tan, and yellow. Assume there are at least 3 of each color. If you pick three candies from a bag, how many color possibilities are there?
(A) 18
(B) 20
(C) 120
(D) 216
(E) 729
2. A code consists of two letters of the alphabet followed by 5 digits. How many such codes are possible?
(A) 7
(B) 10
(C) 128
(D) 20,000
(E) 67,600,000
3. A salad bar has 7 ingredients, excluding the dressing. How many different salads are possible where two salads are different if they don’t include identical ingredients?
(A) 7
(B) 14
(C) 128
(D) 5,040
(E) 823,543
FACTORIAL, PERMUTATIONS, COMBINATIONS
Counting problems like the ones in the last three examples occur frequently enough that they have special designations.
Ordering n Objects (Factorial)
The second example of the multiplication rule asked for the number of ways 6 friends could stand in line. By using the multiplication principle, we found that there were 6 5 4 3 2 1 ways. A special notation for this product is 6! (6 factorial). In general, the number of ways objects can be ordered is n !
Ordering r of n Objects (Permutations)
The third example of the multiplication rule asked for the number of ways you could choose a first (president), second (vice president), and third person (secretary) out of 20 people (r = 3, n = 20). The answer is 20 19 18, or 
. In general, there are 
permutations of r objects of n . This appears as nPr in the calculator menu.
Choosing r of n Objects (Combinations)
In the fourth example of the multiplication rule, we were interested in choosing a committee of 3 where there was no distinction among committee members. Our approach was first to compute the number of ways of choosing officers and then dividing out the number of ways the three officers could hold the different offices. This led to the computation 
. In general, the number of ways of choosing r of n objects is 
. This quantity appears on the calculator menu as nCr. However, there is a special notation for combinations: 
the number of ways r objects can be chosen from n .
Calculator commands for all three of these functions are in the MATH/PRB menu.
These 3 commands can also be found on scientific calculators.
EXERCISES
1. How many 3-person committees can be selected from a fraternity with 25 members?
(A) 15,625
(B) 13,800
(C) 2,300
(D) 75
(E) 8
2. A basketball team has 5 centers, 9 guards, and 13 forwards. Of these, 1 center, 2 guards, and 2 forwards start a game. How many possible starting teams can a coach put on the floor?
(A) 56,160
(B) 14,040
(C) 585
(D) 197
(E) 27
3. Five boys and 6 girls would like to serve on the homecoming court, which will consist of 2 boys and 2 girls. How many different homecoming courts are possible?
(A) 30
(B) 61
(C) 150
(D) 900
(E) 2048
4. In a plane there are 8 points, no three of which are collinear. How many lines do the points determine?
(A) 7
(B) 16
(C) 28
(D) 36
(E) 64
5. If 
, then x =
(A) 0
(B) 1
(C) 4
(D) 5
(E) 10
Answers and Explanations
Venn Diagrams
1. (A) A Venn diagram will help you solve this problem.
The two circles represent males and people who are at most 30 years of age, respectively. The part of the rectangle outside both circles represents people who are in neither category, i.e., females over the age of 30. First fill in the 12 males who are less than or equal to 30 years of age in the intersection of the circles. Since there are 28 males altogether, 16 are male and over 30. Since there are 32 people age 30 or less, there are 20 women that age. Add these together to get 48 people. Since there are 50 in the group, 2 must be women over 30.
Multiplication Rule
1. * (D) There are 6 choices of color for each of the three candies selected. Therefore, there are 6 × 6 × 6 = 216 color possibilities altogether.
2. * (E) The multiplication rule applies. There are (26)(26)(10)5 = 67,600,000 possible codes.
3. * (C) You can either include or exclude each of the seven ingredients in your salad, which means there are 2 choices for each ingredient. According to the multiplication rule, there are 27 = 128 ways of making these yes-no choices.
Factorial, Permutations, Combinations
1. * (C) This is the number of ways 3 objects can be chosen from 25, or 
= 25nCr3 = 2,300.
2. * (B) There are 
ways of choosing the one center, 
ways of choosing the two guards, and 
ways of choosing the two forwards. Therefore, there are 5 × 36 × 78 = 14,040 possible starting teams.
3. * (C) There are 
ways of choosing 2 boys out of 5 and 
ways of choosing 2 girls out of 6. Therefore, there are 10 × 15 = 150 ways of choosing the homecoming court.
4. * (C) Since no three points are collinear, every pair of points determines a distinct line. There are 
such lines.
5. (A) 
for any n.